a: \(5-\dfrac{7}{8}+\dfrac{15}{-20}\)

\(=5-\dfrac{7}{8}-\dfrac{3}{4}\)

\(=\dfrac{5\cdot8-7-3\cdot2}{8}\)

\(=\dfrac{40-7-6}{8}=\dfrac{27}{8}\)

b: \(\dfrac{10}{-13}:\dfrac{-4}{13}\cdot\dfrac{11}{-10}\)

\(=-\dfrac{10}{13}\cdot\dfrac{13}{4}\cdot\dfrac{11}{10}\)

\(=-\dfrac{11}{4}\)

c: \(\dfrac{-3}{2}\cdot\dfrac{5}{2}+\dfrac{3}{-8}\)

\(=\dfrac{-15}{4}+\dfrac{-3}{8}\)

\(=\dfrac{-15\cdot2+\left(-3\right)}{8}=\dfrac{-33}{8}\)

d: \(\dfrac{7}{-8}-\dfrac{-4}{5}:\dfrac{3}{10}\)

\(=-\dfrac{7}{8}+\dfrac{4}{5}\cdot\dfrac{10}{3}\)

\(=-\dfrac{7}{8}+\dfrac{8}{3}\)

\(=\dfrac{-7\cdot3+8\cdot8}{24}=\dfrac{43}{24}\)

e: \(\dfrac{-5}{8}\cdot\dfrac{25}{111}+\dfrac{25}{111}\cdot\dfrac{3}{10}\)

\(=\dfrac{25}{111}\left(-\dfrac{5}{8}+\dfrac{3}{10}\right)\)

\(=\dfrac{25}{111}\cdot\dfrac{-25+12}{40}\)

\(=\dfrac{25}{40}\cdot\dfrac{-13}{111}=\dfrac{-5}{8}\cdot\dfrac{13}{111}=\dfrac{-65}{888}\)

Đặt \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

Thêm bớt \(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\) vào A ta được:

\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)

\(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

Vậy \(\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}}=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}=1\)

\(B=\dfrac{2n-3}{n-2}=\dfrac{2n-4+1}{n-2}=\dfrac{2\left(n-2\right)}{n-2}+\dfrac{1}{n-2}=2+\dfrac{1}{n-2}\)

Do \(2\in Z\Rightarrow B\in Z\) khi \(\dfrac{1}{n-2}\in Z\)

\(\Rightarrow n-2=Ư\left(1\right)\)

\(\Rightarrow n-2=\left\{-1;1\right\}\)

\(\Rightarrow n=\left\{1;3\right\}\)

e/\(E=\dfrac{7}{3}+\dfrac{7}{15}+\dfrac{7}{35}+...+\dfrac{7}{575}\)
\(=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{23.25}\)
\(=\dfrac{7}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{23.25}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{23}-\dfrac{1}{25}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{25}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{24}{25}\)
\(=\dfrac{84}{25}\)
g)\(G=\dfrac{5}{2.4}+\dfrac{5}{4.6}+\dfrac{5}{6.8}+...+\dfrac{5}{100.102}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{100.102}\right)\)
\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+...+\dfrac{1}{100}-\dfrac{1}{102}\right)\)
\(=\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{102}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{25}{51}\)
\(=\dfrac{125}{102}\)
h/\(H=\dfrac{2024}{2.5}+\dfrac{2024}{5.8}+...+\dfrac{2024}{62.65}\)
\(=\dfrac{2024}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{62.65}\right)\)
\(=\dfrac{2024}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{62}-\dfrac{1}{65}\right)\)
\(=\dfrac{2024}{3}\left(\dfrac{1}{2}-\dfrac{1}{65}\right)\)
\(=\dfrac{2024}{3}\cdot\dfrac{63}{130}\)
\(=\dfrac{21252}{65}\)
#TiendatzZz

a/\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{100}{101}\)
\(=\dfrac{50}{101}\)
b/\(B=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}...+\dfrac{4}{49.51}\)
\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(=2\left(1-\dfrac{1}{51}\right)\)
\(=2\cdot\dfrac{50}{51}\)
\(=\dfrac{100}{51}\)
c/\(C=\dfrac{6}{3.5}+\dfrac{6}{5.7}+\dfrac{6}{7.9}+...+\dfrac{6}{99.101}\)
\(=3\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=3\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)
\(=3\cdot\dfrac{98}{303}\)
\(=\dfrac{98}{101}\)
d/\(D=\dfrac{1}{3}+\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{1023}\)
\(=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{31.33}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{31.33}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{33}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{32}{33}\)
\(=\dfrac{16}{33}\)
#TiendatzZz

\(-5,2+3,5.\left(-2\right)-6,9:3+\left(-0,04\right)\\ =-5,2-7-2,3-0,04\\ =-5,2-3-0,04\\ =-8,24\)

TK:

=−5,2−7−2,3−0,04

=−5,2−3−0,04

=−8,24

\(\dfrac{x-3}{-25}=\dfrac{-4}{x-3}\)

\(\Rightarrow\left(x-3\right)\left(x-3\right)=\left(-4\right).\left(-25\right)\)

\(\Rightarrow\left(x-3\right)^2=100\)

\(\Rightarrow\left(x-3\right)^2=10^2\)

\(\Rightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=13\\x=-7\end{matrix}\right.\)

Anh giải bài xong rồi nghỉ sớm ạ! Em chúc anh ngủ ngon ạ! < 3

\(\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}=\)

\(=\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+...++\dfrac{100-99}{99.100}=\)

\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}=\dfrac{1}{3}-\dfrac{1}{100}< \dfrac{1}{3}\)

\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}< \dfrac{1}{3}\)

Đặt số chia là a và thương là b ta có

\(\dfrac{200-13}{a}=b\Rightarrow b=\dfrac{187}{a}\Rightarrow187⋮a\)

\(\Rightarrow a=\left\{1;11;17\right\}\Rightarrow b=\left\{187;17;11\right\}\)